Problem: You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

Solution (in PYTHON3):

Solution:
import math
from collections import Counter

def comb(n,r):
    f=math.factorial
    return f(n)//(f(r)*f(n-r))

N,A,B=list(map(int,input().split()))
v=sorted(list(map(int,input().split())),reverse=True) # 全要素

m=sum(v[:A])/A # 平均値の最大値(A個の場合)

c=Counter(v[:A]) # 大きい方からA個の要素のみ
# この中の最小の要素は，v[A-1]
a=v[A-1]

n=v.count(a) # v内のaの個数
r=c[a] # c内のaの個数

# n個のaの中から，r個を選ぶ方法
ans=comb(n,r)

# n個のaの中から，r個以上min(r+(B-A),n)個以下を選ぶ方法
# ※これはv[0]==v[A]のときのみ
if v[0]==v[A]:
    for i in range(r+1,min(r+(B-A)+1,n+1)):
        ans+=comb(n,i)

print(m)
print(ans)